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Chapter 16: Problem 31

Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid–base properties. a. \(A g_{3} P O_{4}, K_{s p}=1.8 \times 10^{-18}\) b. \(\mathrm{CaCO}_{3}, K_{\mathrm{sp}}=8.7 \times 10^{-9}\) c. $\mathrm{Hg}_{2} \mathrm{Cl}_{2}, K_{\mathrm{sp}}=1.1 \times 10^{-18}\left(\mathrm{Hg}_{2}^{2+} \right.$ is the cation in is the cation in solution.\()\)

Short Answer

Expert verified
The solubility of the compounds is as follows: a. \(Ag_3PO_4\): \(1.32 \times 10^{-5}\, mol/L\) b. \(CaCO_3\): \(2.95 \times 10^{-5}\, mol/L\) c. \(Hg_2Cl_2\): \(3.0 \times 10^{-7}\, mol/L\)

Step by step solution

01

Write the balanced chemical equation and the Ksp expression.

The balanced chemical equation for the dissolution of \(Ag_{3}PO_{4}\) is: \[Ag_3PO_4(s) \rightleftharpoons 3Ag^+(aq) + PO^{3-}_4(aq)\] Now, we can write the Ksp expression as follows: \[K_{sp}= [Ag^+]^3 [PO_4^{3-}]\]
02

Define solubility in terms of molar concentrations and express Ksp in terms of solubility.

Let S be the solubility of \(Ag_3PO_4\) in moles per liter. When 1 mole of \(Ag_3PO_4\) dissolves, it produces 3 moles of \(Ag^+\) and 1 mole of \(PO_4^{3-}\) ions. Therefore, the molar concentrations of the ions in the solution at equilibrium are: \[ [Ag^+] = 3S \] \[ [PO_4^{3-}] = S \] Now we can substitute these expressions into the Ksp expression from step 1: \[1.8 \times 10^{-18} = (3S)^3(S)\]
03

Solve for solubility (S).

Now, we need to solve the equation for S: \[1.8 \times 10^{-18} = 27S^4\] \[S^4 = \frac{1.8 \times 10^{-18}}{27}\] \[S = \sqrt[4]{6.67 \times 10^{-20}}\] \[S = 1.32 \times 10^{-5}\] The solubility of \(Ag_3PO_4\) is approximately \(1.32 \times 10^{-5}\, mol/L\). b. Calculate the solubility of \(CaCO_3\), with \(K_{sp} = 8.7 \times 10^{-9}\).
04

Write the balanced chemical equation and the Ksp expression.

The balanced chemical equation for the dissolution of \(CaCO_3\) is: \[CaCO_3(s) \rightleftharpoons Ca^{2+}(aq) + CO^{2-}_3(aq)\] Now, we can write the Ksp expression as follows: \[K_{sp} = [Ca^{2+}][CO_3^{2-}]\]
05

Define solubility in terms of molar concentrations and express Ksp in terms of solubility.

The molar concentrations of the ions in the solution at equilibrium are: \[ [Ca^{2+}] = S \] \[ [CO_3^{2-}] = S \] Now we can substitute these expressions into the Ksp expression from step 1: \[8.7 \times 10^{-9} = S^2\]
06

Solve for solubility (S).

Now, we need to solve the equation for S: \[S = \sqrt{8.7 \times 10^{-9}}\] \[S = 2.95 \times 10^{-5}\] The solubility of \(CaCO_3\) is approximately \(2.95 \times 10^{-5}\, mol/L\). c. Calculate the solubility of \(Hg_2Cl_2\), with \(K_{sp} = 1.1 \times 10^{-18}\).
07

Write the balanced chemical equation and the Ksp expression.

The balanced chemical equation for the dissolution of \(Hg_2Cl_2\) is: \[Hg_2Cl_2(s) \rightleftharpoons Hg_2^{2+}(aq) + 2Cl^-(aq)\] Now, we can write the Ksp expression as follows: \[K_{sp} = [Hg_2^{2+}][Cl^-]^2\]
08

Define solubility in terms of molar concentrations and express Ksp in terms of solubility.

The molar concentrations of the ions in the solution at equilibrium are: \[ [Hg_2^{2+}] = S \] \[ [Cl^-] = 2S \] Now we can substitute these expressions into the Ksp expression from step 1: \[1.1 \times 10^{-18} = S(2S)^2\]
09

Solve for solubility (S).

Now, we need to solve the equation for S: \[1.1 \times 10^{-18} = 4S^3\] \[S^3 = \frac{1.1 \times 10^{-18}}{4}\] \[S = \sqrt[3]{2.75 \times 10^{-19}}\] \[S = 3.0 \times 10^{-7}\] The solubility of \(Hg_2Cl_2\) is approximately \(3.0 \times 10^{-7}\, mol/L\).

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Most popular questions from this chapter

The solubility of \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\) in a $0.20-M \mathrm{KIO}_{3}\( solution is \)4.4 \times 10^{-8} \mathrm{mol} / \mathrm{L}$ . Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}.\)

Sulfide precipitates are generally grouped as sulfides insoluble in acidic solution and sulfides insoluble in basic solution. Explain why there is a difference between the two groups of sulfide precipitates.

Consider 1.0 L of an aqueous solution that contains 0.10 M sulfuric acid to which 0.30 mole of barium nitrate is added. Assuming no change in volume of the solution, determine the pH, the concentration of barium ions in the final solution, and the mass of solid formed.

Silver cyanide \((\mathrm{AgCN})\) is an insoluble salt with \(K_{\mathrm{sp}}=2.2 \times 10^{-12}\) . Compare the effects on the solubility of silver cyanide by addition of \(\mathrm{HNO}_{3}(a q)\) or by addition of \(\mathrm{NH}_{3}(a q).\)

Magnesium hydroxide, \(\operatorname{Mg}(\mathrm{OH})_{2},\) is the active ingredient in the antacid TUMS and has a \(K_{\mathrm{sp}}\) value of $8.9 \times 10^{-12}\( . If a 10.0 -g sample of \)\mathrm{Mg}(\mathrm{OH})_{2}$ is placed in \(500.0 \mathrm{mL}\) of solution, calculate the moles of OH -ions present. Because the \(K_{\mathrm{sp}}\) value for \(\mathrm{Mg}(\mathrm{OH})_{2}\) is much less than \(1,\) not a lot solid dissolves in solution. Explain how \(\mathrm{Mg}(\mathrm{OH})_{2}\) works to neutralize large amounts of stomach acid.
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